Special Relativity

Philip Harris

University of Sussex

Contents

1 Introduction 4

1.1 Syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Recommended Texts . . . . . . . . . . . . . . . . . . . . . . . 5

2 Background History 6

2.1 The Principle of Relativity . . . . . . . . . . . . . . . . . . . . 7

2.2 Newtonian/Galilean Relativity . . . . . . . . . . . . . . . . . . 9

2.2.1 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2.2 Example II . . . . . . . . . . . . . . . . . . . . . . . . 11

2.2.3 Inverse Transformations . . . . . . . . . . . . . . . . . 12

2.2.4 Measuring Lengths . . . . . . . . . . . . . . . . . . . . 12

2.3 The Clash with Electromagnetism . . . . . . . . . . . . . . . . 14

2.4 The Invention of the Ether . . . . . . . . . . . . . . . . . . . . 15

2.5 The Michelson-Morley Experiment . . . . . . . . . . . . . . . 15

2.6 Frame Dragging and Stellar Aberration . . . . . . . . . . . . . 18

2.7 The Lorentz Transformation . . . . . . . . . . . . . . . . . . . 18

2.8 Poincar�e and Einstein . . . . . . . . . . . . . . . . . . . . . . 20

3 The Lorentz Transformations 21

3.1 Transverse Coordinates . . . . . . . . . . . . . . . . . . . . . . 21

3.2 Time Dilation . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.2.1 Warning . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2.2 The Lifetime of the Muon . . . . . . . . . . . . . . . . 23

3.2.3 The Twins Paradox . . . . . . . . . . . . . . . . . . . . 24

3.3 Lorentz Contraction . . . . . . . . . . . . . . . . . . . . . . . 24

3.3.1 Alternative Approach . . . . . . . . . . . . . . . . . . . 25

3.4 Derivation of Lorentz Transformations . . . . . . . . . . . . . 26

3.4.1 Matrix Formulation . . . . . . . . . . . . . . . . . . . . 29

1

3.5 Help! When Do I Use Which Formula? . . . . . . . . . . . . . 29

3.6 The Doppler Shift . . . . . . . . . . . . . . . . . . . . . . . . . 30

3.7 Synchronisation of Clocks . . . . . . . . . . . . . . . . . . . . 31

3.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

4 Spacetime 34

4.1 Spacetime Events and World Lines . . . . . . . . . . . . . . . 34

4.2 Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.3 Simultaneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

4.4 The Principle of Causality . . . . . . . . . . . . . . . . . . . . 38

4.4.1 Example: Gun�ght on a Train . . . . . . . . . . . . . . 39

4.4.2 Example: A Lighthouse . . . . . . . . . . . . . . . . . 39

4.5 Light Cones . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

4.6 Visual Appearance of Moving Objects . . . . . . . . . . . . . 40

4.7 Oblique Axes . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.8 More Paradoxes . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.8.1 The Pole and Barn Paradox . . . . . . . . . . . . . . . 42

4.8.2 The Thin Man and the Grid . . . . . . . . . . . . . . . 42

4.8.3 Twins Paradox Revisited . . . . . . . . . . . . . . . . . 43

5 Dynamics and Kinematics 50

5.1 Relative Velocities . . . . . . . . . . . . . . . . . . . . . . . . 50

5.2 Lorentz Transformation of Velocities . . . . . . . . . . . . . . 50

5.3 Successive Lorentz Transformations . . . . . . . . . . . . . . . 52

5.4 Velocity Parameter . . . . . . . . . . . . . . . . . . . . . . . . 53

5.5 Relativistic Dynamics . . . . . . . . . . . . . . . . . . . . . . . 53

5.6 Consequences . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

5.7 Lorentz Transformation of Energy-Momentum . . . . . . . . . 58

5.7.1 The DiÆcult Way . . . . . . . . . . . . . . . . . . . . . 58

5.7.2 The Easy Way . . . . . . . . . . . . . . . . . . . . . . 62

5.7.3 Matrix Notation . . . . . . . . . . . . . . . . . . . . . 64

5.8 Example: Collision Threshold Energies . . . . . . . . . . . . . 64

5.9 Kinematics: Hints for Problem Solving . . . . . . . . . . . . . 65

5.10 Doppler Shift Revisited . . . . . . . . . . . . . . . . . . . . . . 66

5.11 Aberration of Light Revisited . . . . . . . . . . . . . . . . . . 67

5.12 Natural Units . . . . . . . . . . . . . . . . . . . . . . . . . . . 68

2

6 Four-Vectors 70

6.1 Scalar Products . . . . . . . . . . . . . . . . . . . . . . . . . . 71

6.2 Energy-Momentum Four-Vector . . . . . . . . . . . . . . . . . 72

6.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6.3.1 Collision Threshold Energies . . . . . . . . . . . . . . . 74

6.3.2 Matter-Antimatter annihilation . . . . . . . . . . . . . 75

7 Relativity and Electromagnetism 78

7.1 Magnetic Field due to a Current . . . . . . . . . . . . . . . . . 78

7.2 Lorentz Transformation of Electromagnetic Fields . . . . . . . 82

7.3 Electromagnetic Waves . . . . . . . . . . . . . . . . . . . . . . 82

7.3.1 Phase . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

7.3.2 Wave four-vector . . . . . . . . . . . . . . . . . . . . . 84

7.3.3 Doppler Shift . . . . . . . . . . . . . . . . . . . . . . . 84

3

Chapter 1

Introduction

1.1 Syllabus

� The principle of relativity; its importance and universal application.

� Revision: Inertial frames and transformations between them. Newton's

laws in inertial frames.

� Acquaintance with historical problems of con ict between electromag-

netism and relativity.

� Solution?: The idea of ether and attempts to detect it. Michelson-

Morley experiment - observed constancy of speed of light.

� Einstein's postulates.

� Lorentz transformations; Lorentz contraction and time dilation.

� Examples: Cosmic ray experiments (decay of mu meson).

� Transformation of velocities.

� Familiarity with spacetime (Minkowski) diagrams, intervals, causality.

� Visual appearance of moving objects (not required for exam).

� Failure of simultaneity at a distance.

� Resolution of so-called \paradoxes" in relativity.

4

� Transformation of momentum and energy. Mass-energy relations. En-

ergy/momentum invariant. Collisions. Creation of particles.

� Doppler e�ect.

� Acquaintance with four-vectors (not required for exam).

� Introductory acquaintance with relativistic e�ects in electromagnetism:

magnetic force due to current-bearing wire.

1.2 Recommended Texts

� Special Relativity, A.P. French, pub. Chapman and Hall, ISBN 0412343207.

A \standard" text since 1971.

� The Feynman Lectures on Physics, Vol. I chaps. 15-17; Vol II sections

13.6, 13.7 and chapter 42. The classic introduction to all branches of

physics; brilliant as ever! Perhaps a little demanding to begin with,

but well worth going back to read later in the course.

� Relativity Physics, R.E. Turner, pub. Routledge and Kegan Paul, 1984;

ISBN 0-7102-0001-3. Out of print. Straightforward approach, probably

one of the simplest texts for beginners in the �eld. This is the text I

would recommend for this course if it were in print.

� Special Relativity, J.G. Taylor, pub. Oxford, 1975. Out of print; try

the library. Simple and straightforward approach.

� Spacetime Physics, Taylor and Wheeler . An alternative presentation,

based entirely on geometric spacetime transformations; full of excellent

examples. Out of print; try the library.

� The Special Theory of Relativity, D. Bohm, pub. Routledge, ISBN

0-415-14809-X. Conceptural structure and underlying physical ideas

explored thoroughly and clearly, but perhaps not for the beginner.

This set of lecture notes is based principally upon material drawn from

these sources.

5

Chapter 2

Background History

Einstein's theory of relativity has a formidable reputation as being incredibly

complicated and impossible to understand. It's not! The principle of relativ-

ity itself, the single, simple idea upon which Einstein's theory is based, has

been around since the time of Galileo. As we shall see, when it is applied to

objects that are moving extremely fast, the consequences seem strange to us

because they are outside our everyday experience; but the results make sense

and are all self-consistent when we think about them carefully. We can sum-

marize the major corrections that we need to make to Newton's equations

of motion as follows: Firstly, when an object is in motion, its momentum p

is larger than expected, its length l shrinks in the direction of motion, and

time t slows down, in each case by a factor

=1p

1� v2=c2; (2.1)

where v is the velocity of the object and c is the speed of light:

c = 299792458m=s(exactly; byde�nition)

= 186; 282miles=second

= 30cm=ns; inunitswecangrasp:

This seems confusing at �rst because we are used to assuming, for example,

that the length l of any object should be constant. For everyday purposes,

the correction is tiny | consider the International Space Station, moving

at 8 km/s in orbit; its length is about 1 part per billion less than if it were

at rest, and time on board moves more slowly by the same factor. But

6

for many particles moving near the speed of light, the fact that time slows

down (and hence lifetimes are longer) with velocity by the factor (2.1) has

been well veri�ed. The same applies to the increasing momentum { which

demonstrates immediately the well-known principle that one can never push

an object hard enough to accelerate it to the speed of light, since, as it goes

faster and faster, you have to push harder and harder to obtain a given

increase in velocity. Loosely speaking, it acts as though the mass increases

with velocity.

Secondly, as we shall discover about halfway through the course, there

emerges naturally what may well be the most famous equation in the world:

E = mc2: (2.2)

These formulae were not found experimentally, but theoretically, as we

shall see.

Einstein's 1905 relativity paper, \On the Electrodynamics of Moving Bod-

ies", was one of three he published that year, at age 26, during his spare time;

he was at the time working as a patent clerk in Zurich. Another was a pa-

per explaining Brownian motion in terms of kinetic theory (at a time when

some people still doubted the existence of atoms), and the third proposed

the existence of photons, thus laying the foundations for quantum theory and

earning him the Nobel prize (relativity being too controversial then).

Einstein wrote two theories of relativity; the 1905 work is known as \spe-

cial relativity" because it deals only with the special case of uniform (i.e.

non-accelerating) motion. In 1915 he published his \general theory of rel-

ativity", dealing with gravity and acceleration. Strange things happen in

accelerating frames; objects appear to start moving without anything push-

ing them... During this course we shall only deal with special relativity.

2.1 The Principle of Relativity

As we use our telescopes to look ever farther out into the universe, some

relevant questions present themselves:

� Is space homogeneous? I.e., is it the same everywhere | are the laws

of physics the same in distant galaxies as they are here on Earth?

7

� Is it isotropic | is it the same in all directions, or is there some de�ning

\axis" or direction that is preferred in some way? Is, for example, the

speed of light the same in all directions?

� Are the laws of physics constant in time?

� And �nally, are the laws of physics independent of uniform relative

motion?

By looking at light from the most distant visible galaxies, more than 10

billion light years away, we can recognise the spectra of hydrogen atoms. As

far as we can tell, those hydrogen atoms are the same everywhere. And be-

cause the light was emitted so long ago, it seems clear that the laws of physics

are indeed constant in time (with the possible exception of the gravitational

constant G; its rate of change is very diÆcult to measure, but no variation

has been seen so far).

The last of these questions lies at the core of relativity. If I perform

an experiment on board a rocket that is moving uniformly through space

(remember, we aren't dealing with acceleration or gravity here), will I get

the same result as somebody doing the same experiment on another rocket

moving at a di�erent speed? \Common sense" suggests that there should be

no di�erence. This \common sense" idea is known in physics as the principle

of relativity, and it was �rst proposed by Galileo. Here is Newton's de�nitive

statement of it as a corollary to his laws of motion:

\The motions of bodies included in a given space are the same

among themselves, whether that space is at rest or moving uni-

formly forward in a straight line."

Meaning: if a spaceship is drifting along at a uniform velocity, all exper-

iments and phenomena inside the spaceship will be just the same as if the

ship were not moving. There is no \preferred" inertial (i.e. non-accelerating)

frame of reference which is \at rest" in the universe; and therefore, you can-

not tell how \fast" a spaceship, or car, or whatever, is moving by doing

experiments inside | you have to look outside to compare, in order to see

how fast it is moving relative to its surroundings.

Galileo considered ordinary ships instead of spaceships: he pointed out

that a rock dropped from the top of the mast will hit the same spot on deck

whether the ship is stationary or moving along uniformly.

8

This is a simple and appealing idea which, of course, needed to be tested

experimentally. Before we go any further, though, let us familiarise ourselves

with the meaning of relativity in the everyday world of Newtonian mechanics.

2.2 Newtonian/Galilean Relativity

Consider two people, Tony (standing still) and Bill (walking past at velocity

u). Tony has a \reference frame" S in which he measures the distance to a

point on the pavement ahead of him, and calls it x. Bill, who walks past at

time t = 0, has a \reference frame" S 0 in which he measures the (continually

changing) distance to the same point, and calls it x0. Then, a Galilean

transformation links the two frames:

x0 = x� ut

y0 = y

z0 = z

t0 = t: (2.3)

This is commonsense: we can see how the distance that Bill measures to the

point decreases with time, until it goes negative when Bill actually walks

past the point.

What about velocities? Suppose Tony is standing still to watch a bird y

past at speed v. He now calls the (changing) distance to the bird x, and

v =dx

dt:

If we di�erentiate 2.3, we see that, according to Bill, the bird is ying past

with speed

v0 =

dx0

dt0=

dx0

dt

=dx

dt� u

= v � u: (2.4)

We see too that acceleration is the same in both frames,

a0 =

dv0

dt0=

dv

dt= a; (2.5)

so Newton's law, F = ma, will be the same in both frames of reference;

likewise conservation of momentum holds true in both frames.

9

x′x

y′y

ut

P(x,y,z) or

(x′,y′,z′)

BillTony

Figure 2.1: A pair of coordinate systems in relative motion.

2.2.1 Example

Suppose Tony is standing by the railway tracks, watching a train go past

to the east at 25 m/s. At the same time, a plane is ying overhead (again

eastwards) at 200 m/s. Meanwhile, a car drives away to the north at 25 m/s.

What does the scene look like to Bill, who is sitting on the train?

We have to remember here that velocity is a vector. In order to transform

from Tony's frame of reference to Bill's, we will have to use a vector version

of (2.4):

v0 = v � u: (2.6)

As above, u is the velocity of Bill relative to Tony. Let's call Eastwards

the i direction and northwards the j direction. Then

u = 25i:

The velocity of the plane is, according to Tony,

v = 200i:

Therefore, the velocity of the plane relative to Bill is

v0 = v � u

10

= 200i� 25i = 175i:

As seen from the train, then, the plane is ying past eastwards at a speed of

175 m/s.

Tony calculates the velocity of the car to be

v =25j:

Therefore, from Bill's frame of reference, the car is moving with velocity

v0 = 25j� 25i:

Thus, the car is moving in a northwesterly direction relative to the train.

2.2.2 Example II

Tony is playing snooker. The white ball, which has mass m and moves with

velocity

v =13icm=s;

hits a stationary red ball, also of mass m, in an elastic collision. The white

ball leaves the collision with velocity vw = 11:1i+4:6j (i.e. at 12 cm/s at an

angle of 22.6Æ above the horizontal), and the red ball leaves at a velocity of

vr = 1:9i� 4:6j (which is 5 cm/s at an angle of 67.4Æ below the horizontal).

You can check that momentum and energy are conserved. Suppose now that

Bill is walking past with a velocity of u = 13i. What does the collision look

like to him?

We know straight away that, since he is moving with the same speed as

the white ball had initially, it is at rest in his reference frame; this of course

agrees with equation (2.6). What about the red ball before the collision? In

his frame, it is no longer at rest; instead, it is moving \backwards", with

velocity 0� 13i = �13i: After the collision, we obtain for the white ball

v0

w= (11:1i+ 4:6j)� 13i

= �1:9i+ 4:6j;

whereas the red ball moves with velocity

v0

r= (1:9i� 4:6j)� 13i

= �11:1i� 4:6j:

11

From Bill's point of view, then, the collision is essentially a mirror image

of the collision as seen from Tony's reference frame. For Bill, it is the red ball

that moves in and hits the stationary white ball. As expected, momentum

and energy are conserved in the two frames.

2.2.3 Inverse Transformations

Naturally, if we have a set of coordinates in Bill's frame of reference and

we want to know how they look from Tony's point of view, we just need to

realise that, according to Bill, Tony is moving past with a velocity of �u,and so the (inverse) transformation is

x = x0 + ut

0

y = y0

z = z0

t = t0: (2.7)

2.2.4 Measuring Lengths

Tony is sitting on a train, which is moving at speed u. He has paced the

corridors from one end to the other, and calculates that its length is 100 m.

Bill, meanwhile, is standing by the tracks outside, and is curious to calculate

for himself how long the train is. Several ways suggest themselves.

1. He can note where he is standing when the front of the train passes,

then run towards the back end and note where he is standing when the

back end passes, and subtract the two distances. This will obviously

give the wrong answer.

2. He can set up a row of cameras, which will take their pictures simulta-

neously. The separations between the cameras that see the front and

the rear of the train gives the length of the train.

3. He can stand still, and time the train going past. Using his Doppler

radar gun he measures the speed of the train, and from the speed and

the time he calculates the length.

4. He can do some mixture of these, measuring the position of the front

of the train at one time and the back of the train at another, and

12

compensate for the train's velocity by calculating where the two ends

would be at some particular moment in time, t0 = 0.

Looking at this formally, suppose that, according to Tony, the front of the

train is at x2 and the rear is at x1, so the length is (x2 � x1). Bill measures

x01 and x

02 at times t

01, t

02. He calculates that at time t

0 = 0, the front of the

train was at position

x02 � ut

02;

and the rear was at position

x01 � ut

01:

The length of the train is therefore given by the di�erence between these

positions:

x02 � ut

02 � x

01 + ut

01:

In using equation (2.3) to transform from Tony's to Bill's frame, we remember

that Bill is moving past with a velocity of �u, so

x01 = x1 + ut1

x02 = x2 + ut2

t01 = t1

t02 = t2:

Therefore, the length will be

(x02 � x01)� u (t02 � t

01) = (x2 � x1) ; (2.8)

which agrees with Tony's length measurement. Comparing with the options

above, we have:

1. corresponds to forgetting that t02 6= t01, just using x

02 � x

01; and getting

the answer wrong.

2. corresponds to measuring the coordinates of the two ends at the same

time; t02 = t01, so x

02 � x

01 = (x2 � x1).

3. corresponds to measuring at the same place; x02 = x01, so the length is

u (t01 � t02) :

4. corresponds to using (2.8) to compensate for the speed of the train.

This is just a matter of putting commonsense on a �rm footing.

13

2.3 The Clash with Electromagnetism

Newton's laws reigned supreme in mechanics for more than 200 years. How-

ever, diÆculties arose in the mid-19th century with studies of electromag-

netism. All electrical and magnetic e�ects could be summarised nicely in

Maxwell's Laws, which we won't go into here. The problem was that, unlike

Newton's laws, these were not invariant under a Galilean transformation |

the principle of relativity didn't seem to be valid for electricity or magnetism!

Therefore, in a moving spaceship, it seemed that electromagnetic (including

optical) phenomena would be di�erent than they would be in a laboratory

that was \at rest", and one should be able to determine the speed of the

spaceship by doing optical or electrical experiments. Can you imagine, for

example, if all of the magnets on the space shuttle were to get weaker in the

�rst half of its orbit and stronger again in the second half, just because it was

moving in di�erent directions through space as it went around the Earth?

Something was wrong!

In particular, the con ict became apparent where Maxwell's Laws pre-

dicted a constant speed of light, independent of the speed of the source. Sound

is like this; it moves through the air at the same speed, regardless of the speed

of the source. (The speed of the source changes the frequency, or pitch, of

the sound, via the famous Doppler e�ect, but not its speed). As an example,

suppose that Tony is standing still, on a calm day, and Bill is paragliding

past at 30 m/s. Sound from the rear moves past Tony (in the same direc-

tion as Bill is going) at 330 m/s. The apparent speed of the sound wave, as

measured by Bill, is

v0 = v � u = 300m=s:

So, Bill can measure this speed, and deduce that he is moving at 30 m/s |

relative to the air...

Suppose now Bill is on a spaceship, moving at u = 2 � 108 m/s, and

is overtaken by light moving at c = 3 � 108 m/s. By measuring the speed

of light going past, can Bill measure the speed of the spaceship? Newton's

laws, using the Galilean transformation, would suggest that Bill would see

the light going past at 1 � 108 m/s, from which he could deduce his speed;

but Maxwell's laws, which predict that the light would pass Bill at 3 � 108

m/s regardless of his speed, clearly disagree.

14

2.4 The Invention of the Ether

Since, by the 1870s, Newton's Laws had stood the test of time for two cen-

turies, and Maxwell's Laws, having a vintage of just 20 years or so, were

young upstarts, the natural assumption was that Maxwell's Laws needed

some modi�cation. The �rst obvious conclusion was that, just as sound

needed a medium to travel through | and the speed was constant relative

to the medium | so light must need a medium too. (Remember that no-

body had come across the idea of a wave without a medium until then). The

Victorian scientists named this medium the ether (or �ther, if you prefer).

It had to have some strange properties:

� Invisibility, of course.

� It was massless.

� It �lled all of space.

� High rigidity, so light could travel so quickly through it. (Something

that springs back fast carries waves more quickly than something soft;sound

travels faster through iron than air).

� It had no drag on objects moving through it: the Earth isn't slowed

down in its orbit.

� Other curious properties had to be assumed to explain new experimen-

tal results, such as...

2.5 The Michelson-Morley Experiment

If the Earth is really a \spaceship" moving through the ether, the speed

of light in the direction of Earth's motion should be lower than it is in a

direction at right angles to this. By measuring these speeds, we should

therefore be able to detect Earth's absolute velocity relative to the ether.

The most famous experiment that tried to do this was the Michelson-Morley

experiment, in 1887. Here is how it works (see diagram):

There is a light source at A. A glass plate at B is half-silvered, so half

of the light is re ected up to C (ignore the dashed lines for now), where it

hits a mirror and comes back down. The other half of the light carries on

15

E′

B′

u

Wavesout

Waves inphase

E

B

C′C

Source A

Figure 2.2: The Michelson-Morley experiment.

through to E, where it also is re ected back. The two beams are recombined

on the other side of B, where they make interference fringes (bright where

crest meets crest, dark where crest meets trough).

Now, suppose the apparatus starts to move to the right at velocity u.

Suppose the time now needed for the light to go from B to E is t1. In this

time, mirror E has travelled distance ut1 to the right, so the total distance

the light has to go is

L+ ut1 = ct1;

since the light is travelling at speed c.

When it bounces back, B is moving in to meet it, so it has a shorter

distance to go; if it takes time t2; we have

L� ut2 = ct2:

Rearranging these equations,

t1 =L

(c� u)

t2 =L

(c+ u)

16

we �nd that the total time for the round trip is

t1 + t2 =L (c+ u) + L (c� u)

(c� u) (c + u)

=2Lc

(c2 � u2)

=2L=c

(1� u2=c2): (2.9)

Now let's do the same calculation for the light that bounces o� C. The

ray follows the hypotenuse of a triangle, and so travels the same distance in

each leg. If each leg takes time t3; and is therefore a distance ct3, we have

(ct3)2= L

2 + (ut3)2;

or

c2t2

3 � u2t2

3 = L2

t2

3

�c2 � u

2�

= L2

so

t3 =Lp

(c2 � u2):

Since the return trip is the same length, the total round trip takes a time of

2t3 =2L=cp

(1� u2=c2): (2.10)

So, the times taken to do the two round trips are not the same.

In fact, the lengths of the arms L cannot be made exactly the same. But

that doesn't matter: what we have to do is to rotate the interferometer by

90Æ, and look for a shift in the interference fringes as we move through the

ether in the direction of �rst one, and then the other arm.

The orbital speed of the Earth is about 30 km/s. Any motion through

the ether should be at least that much at some time of the night or day at

some time of the year | but nothing showed up! The velocity of the Earth

through the ether could not be detected.

17

2.6 Frame Dragging and Stellar Aberration

When an aeroplane ies through the air, or a ship moves through the water, it

drags a \boundary layer" of uid along with it. Was it possible that the Earth

in its orbit was somehow \dragging" some ether along? This idea | known

as \frame dragging" | would explain why Michelson and Morley could not

�nd any motion of the Earth relative to the ether. But this had already

been disproved, by the phenomenon known as stellar aberration, discovered

by Bradley in 1725.

Imagine a telescope, on a \still" Earth, pointed (for simplicity) to look

at a star vertically above it. (See diagram). Now, suppose that the Earth is

moving (and the telescope with it), at a speed v as shown. In order for light

that gets into the top of the telescope to pass all the way down the moving

tube to reach the bottom, the telescope has to be tilted by a small angle Æ,

where

tan Æ =v

c:

(In fact, as we shall see, the angle is actually given by

tan Æ =v=cp

1� v2=c2;

but the correction factor is tiny). The angle Æ oscillates with Earth's orbit,

and, as measured, agrees with Earth's orbital speed of about 30 km/s. In

this case, the ether cannot be being dragged along by the Earth after all,

otherwise there would be no aberration.

2.7 The Lorentz Transformation

Notice that the di�erence in travel times for the two arms of the Michelson

interferometer is a factor of

1p(1� u2=c2)

:

Lorentz suggested that, because all of the electrons in all of the materials

making up the interferometer (and everything else) should have to interact

with the ether, moving through the ether might make materials contract by

just this amount in the direction of motion, but not in transverse directions.

18

(b)(a)

v

Figure 2.3: The aberration of starlight. (a) Stationary telescope. (b) Moving

telescope.

In that case, the experiment would give a null result! (Fitzgerald had also

noticed that this \�x" would work, but could not suggest what might cause

it). In developing this idea further, Lorentz found that clocks that were

moving through the ether should run slowly too, by the same amount. He

noticed that if, instead of the Galilean transformations 2.3, he made the

substitutions

x0 =

x� utp1� u2=c2

(2.11)

y0 = y

z0 = z

t0 =

t� ux=c2p

1� u2=c2:

into the Maxwell equations, they became invariant! These equations are

known as a Lorentz transformation.

19

2.8 Poincar�e and Einstein

To everyone else, naturally, Lorentz's solution looked like an arti�cial fudge,

invented just to solve this problem. People continued to try to discover an

\ether wind", and every time an explanation had to be made up as to why

nature was \conspiring" to thwart these measurements. In the end, Poincar�e

pointed out that a complete conspiracy is itself a law of nature; he proposed

that there is such a law of nature, and that it is not possible to discover

an ether wind by any experiment; that is, there is no way to determine an

absolute velocity. Poincar�e's principle of relativity (1904) states that:

The laws of physics should be the same in all reference frames

which move in uniform motion with respect to one another.

To this, Einstein added his second postulate:

The velocity of light in empty space is the same in all reference

frames, and is independent of the motion of the emitting body.

The second postulate as stated is not very general, and it implies that there

is something special about the behaviour of light. This is not the case at all.

We should restate it as:

There is a �nite speed that is the same relative to all frames of

reference.

The fact that light in vacuo happens to travel at that speed is a consequence

of the laws of electromagnetism { and of the �rst postulate. Let us re-

emphasize: there is nothing particularly special about light that makes it

somehow magically change the properties of the universe. It is the fact that

there is a �nite speed that is the same for all observers that runs counter to

our instincts, and that has such interesting consequences.

We will from now on assume that these postulates are true, and see what

experimental results we can predict from them. In fact, as we stated earlier,

relativity has passed every experimental test that has ever been proposed for

it. It is now so deeply ingrained in our thinking that, unlike other \laws"

or \theories" that apply to particular branches of physics, relativity is used

as a sort of check | any theory that we produce has to be consistent with

relativity or else it is at best an approximation.

20

Chapter 3

The Lorentz Transformations

3.1 Transverse Coordinates

Let us take two rulers that are exactly the same, and give one to a friend

who agrees to mount it on his spaceship and y past us at high speed in the

x direction. His ruler will be mounted in the transverse (y) direction. As he

ies by, we hold pieces of chalk at the 0 and 1 m marks on our ruler, and

hold it out so they make marks on his ruler. When he comes back, we look

to see where those marks are. What do we �nd? They must, of course, be

one metre apart, because if they were di�erent then we would have a way of

knowing which of us was \really" moving. Thus, for transverse coordinates,

the transformations between his frame and ours must be

y0 = y;

z0 = z:

3.2 Time Dilation

Imagine a simple clock, a \light clock", consisting of light bouncing between

two mirrors separated by a distance l (see diagram). Each time the light hits

one of the mirrors, the clock gives out a \tick". Let us make a pair of these,

and give one to our friend to take abord his spaceship, while we keep the

other on Earth. The clock on the spaceship is mounted perpendicular to the

direction of motion, just like the ruler that we used last time.

As our friend ies past, we watch the light bouncing between the mirrors.

21

(b)

v

(a)

Figure 3.1: A \light clock", as seen in its rest frame (a) and from a frame

(b) in which it is moving with velocity v.

But to us, instead of just going up and down, the light makes a zigzag motion,

which means that it has to go further. Between \ticks", therefore, whereas

the light in our clock covers a distance l in time

t = l=c;

the light in the clock on board the spaceship covers a distancepl2 + v2t02;

which it must do in time

t0 =

pl2 + v2t02

c:

Substituting for l, we obtain

t0 =

pc2t2 + v2t02

c

) t02 = t

2 +�v2=c

2�t02

) t2 =

�1� v

2=c

2�t02

22

and therefore

t = t0p1� v2=c2: (3.1)

So, if it takes time t for our clock to make a \tick", it takes time t0 =

t=p1� v2=c2 | which is always greater than t | for the clock in the space-

ship to make a tick. In other words, it looks to us as though the moving clock

runs slowly by a factorp1� v2=c2. Is this just an illusion | something to

do with the type of clock? No: suppose instead we had a pair of another

type of clock (mechanical, quartz, a \biological clock", whatever), that we

agree keeps time with our simple \light clock". We keep one on Earth, and

check that it keeps time. Our friend takes the other one along; but if he

notices any discrepancy between the clocks, then we have a way to tell who

is \really" moving | and that is not allowed!

Furthermore, while it appears to us that his clocks run slowly, it appears

to him that our clocks also run slowly, by exactly the same reasoning! Each

sees the other clock as running more slowly.

A time span as measured by a clock in its own rest frame is called the

proper time. This is not meant to imply that there is anything wrong with

the measurement made from another frame, of course.

3.2.1 Warning

You should be very careful when you use formula 3.1 | it is very easy

to become confused and to use it the wrong way around, thus contracting

instead of dilating time!

3.2.2 The Lifetime of the Muon

How can we test this, without having any extremely fast spaceships handy?

One early test was provided by looking at the lifetime of a particle called the

mu-meson, or muon. These are created in cosmic rays high in the atmosphere,

and they decay spontaneously after an average of about 2.2 �10�6 s; thus,even travelling close to the speed of light, they should not be able to travel

more than about 600 m. But because they are moving so close to the speed

of light, their lifetime (as measured on the Earth) is \dilated" according to

equation (3.1), and they live long enough to reach the surface of the Earth,

some 10 km below... Muons have been created in particle accelerators, and

their lifetimes measured as a function of their speed; the values are always

seen to agree with the formula.

23

3.2.3 The Twins Paradox

The most famous apparent \paradox" in relativity concerns two twins, Peter

and Paul. When they are old enough to drive spaceships, Paul ies away at

very high speed. Peter, who stays on Earth, watches Paul's clocks slowing

down; he walks and talks and eats and drinks more slowly, his heart beats

more slowly, and he grows older more slowly. Just as the muons lived longer

because they were moving, so Paul lasts longer too. When, in the end, he

gets tired of travelling around and comes back to Earth to settle down, he

�nds that he is now younger than Peter! Of course, according to Paul, it is

Peter who has moved away and come back again | so shouldn't Peter be

the younger one?

In fact, the two reference frames are not equivalent. Paul's reference

frame has been accelerating, and he knows this; he is pushed towards the

rear of the spaceship as it speeds up, and towards its front as it slows down.

There is an absolute di�erence between the frames, and it is clear that Paul

is the one who has been moving; he really does end up younger than Peter.

3.3 Lorentz Contraction

Consider one of the apparently long-lived muons coming down through the

atmosphere. In its rest frame, it is created at some time t = 0, at (let us

say) the origin of coordinates, x = 0. At some later time t (but at the same

spatial coordinate x = 0), the surface of the Earth moves up rather quickly

to meet it. If, say, t = 10�6 s, and v = 0:995c, the length of atmosphere that

has moved past it is vt = 300 m. But this is far less than the 10 km distance

separating the events in Earth's frame! It seems that not just time, but also

length is changed by relative motion.

To quantify this, let the distance that the muon travels in Earth's frame

be x0 = vt0; note that, from the muon's point of view, it is the Earth that is

moving, and so we use primes to denote Earth's reference frame. The length

of atmosphere moving past the muon in its rest frame is x = vt: Thus,

x = vt = vt0p1� v2=c2

= x0p1� v2=c2: (3.2)

Therefore, the length of atmosphere x as measured by the muon is contracted

by a factorp1� v2=c2 relative to the length measured on Earth. Likewise,

24

lengths in the muon's rest frame are contracted as seen by Earth-bound

observers; but, because the muon is a point-like particle, such lengths are

diÆcult for us to measure directly.

3.3.1 Alternative Approach

We can work this out again from scratch by mounting our spaceship clock

in the direction of motion (see diagram). The light leaving the rear mirror

is now \chasing" the front mirror, and has to move further to meet it than

it would if the clock were stationary. The distance between the mirrors is l

for our stationary clock on Earth; let's call it l0 for the moving clock. If the

time (as seen from Earth) for the light to travel from the rear to the front

mirror is t01; then

ct01 = l

0 + vt01:

x

time

v∆t1

v∆t2

Figure 3.2: A light clock moving parallel to its axis. Note the vertical axis

here represents time.

For the return journey, where the rear mirror moves up to meet the light,

the distance is shorter:

ct02 = l

0 � vt02:

25

Therefore,

t01 =

l0

c� v;

t02 =

l0

c+ v

and the total time for the light to travel in both directions is

t0 = t

01 + t

02 =

l0 (c+ v) + l

0 (c� v)

(c� v) (c+ v)

=2l0c

c2 � v2:

Now, for the stationary clock the total \there-and-back" time is

t =2l

c:

But we already know that the moving clock is running slowly;

t0 =

tp1� v2=c2

:

Therefore, the total time

2l0c

c2 � v2= t

0 =tp

1� v2=c2

=2l

cp1� v2=c2

;

and so

l0 = l

p1� v2=c2:

This shows us that lengths are actually contracted in the direction of motion.

3.4 Derivation of Lorentz Transformations

We have seen that lengths and times are both modi�ed when bodies are in

motion. We now derive the Lorentz transformations; this involves just a little

algebra; the procedure is entirely based on Einstein's two postulates.

26

Let us start with a fairly general set of linear transformations:

x0 = ax� bt

t0 = dx+ et:

The distances and times correspond to measurements in reference frames

S; S0.We have here assumed

� common origins (x0 = 0 at x = 0) at time t = t0 = 0:

� linear transformations; there are no terms in, e.g., x2. To see that this

is reasonable, consider the point x0 = 0, which is the origin of the S 0

frame. It is moving with velocity

v = dx=dt = b=a = constant;

if there were x2 terms present, the speed would depend on position,

and the velocity would not be uniform.

� Now, the origin of S (i.e., x = 0) moves with velocity �v in the S0

frame, so for x = 0;

x0 = �bt; t0 = et

) dx0

dt0= �b

e= �v:

Therefore, b = ev; but since we have b = av; we know that e = a:

So far, then,

x0 = ax� avt

t0 = dx + at:

� Now we require light to travel with velocity c in all frames, in other

words x = ct is equivalent to x0 = ct0:

ct0 = a:ct� avt

t0 = d:ct+ at:

27

Substitute for t0:

dc2t+ act = act� avt:

Therefore,

dc = �av=c;giving

x0 = a (x� vt) (3.3)

t0 = a

�t� vx=c

2�:

The constant a we can guess from our previous look at time dilation and

space contraction... but here we derive it by requiring symmetry between

the observers. Suppose that we are sitting in the S0 frame, looking at the

S frame. We certainly expect the same relation to hold true the other way

around, if we swap

x ! x0

t ! t0;

but we have to remember that the velocity is reversed:

v ! �v:

So,

x = a (x0 + vt0) (3.4)

t = a�t0 + vx

0=c

2�:

As both 3.3 and 3.4 must hold simultaneously,

x0 = a

�a (x0 + vt

0)� va�t0 + vx

0=c

2�

= a2x0�1� v

2=c

2�

so

a =1p

1� v2=c2:

� We normally give a the symbol :

=1p

1� v2=c2;

28

� ...and de�ne � = v=c, so

=1p

1� �2:

The complete Lorentz transformations are therefore

x0 = (x� �:ct) (3.5)

y0 = y

z0 = z

ct0 = (ct� �x) :

The inverse transformations are then

x = (x0 + �:ct0)

ct = (ct0 + �x0) ;

with the y and z coordinates remaining una�ected as before. Using ct instead

of just t gives all of the variables the dimensions of distance, and displays

the implicit symmetry between the �rst and last transformation equations.

Notice, as usual, the limiting speed of light: if one frame moves faster

than light with respect to another, becomes imaginary, and one or other

frame would then have to have imaginary coordinates x0; t0:

3.4.1 Matrix Formulation

It is clear that (3.5) can be written as0BB@x0

y0

z0

ct0

1CCA =

0BB@ 0 0 �� 0 1 0 0

0 0 1 0

�� 0 0

1CCA0BB@

x

y

z

ct

1CCA :

3.5 Help! When Do I Use Which Formula?

A little thought will tell you when you can use the simple Lorentz contraction

/ time dilation formulae, and when you must use the full Lorentz transfor-

mations.

29

We saw earlier that, if Bill wants to calculate the length of Tony's train

as it passes, he can either measure both ends simultaneously or else make

separate measurements but allow for the distance the train has moved in the

time between measurements. However, when we specify events, the distance

between them depends upon the frame of reference even in Galilean relativity.

For example, a 100 m long train is moving at 50 m/s. At t = 0, the driver

at the front spills his co�ee; one second later, the guard at the back of the

train drops his sandwich. From Tony's point of view, these events took place

100 m apart; but in Bill's frame of reference, they are only separated by 50

m. The Galilean transformations take care of the distance the guard travels

forwards during the 1 s between the events. The same principle applies in

Einstein's relativity, of course.

Basically, then, if you are concerned with objects | rulers, rockets, galax-

ies | you can calculate the lengths in the various frames of reference just by

using the Lorentz contraction. Likewise, if you are concerned with the time

that has passed on a clock (atomic, biological, whatever) that is at rest in one

particular frame of reference, you can safely use the time dilation formula.

But, if you are considering separate events, each with their own space and

time coordinates, you must use the full Lorentz transformations if you want

to get the right answer.

3.6 The Doppler Shift

Consider a source of light of a given frequency �, at rest. Every t = 1=�

seconds it emits a new wavefront.

Now let the source move slowly towards us (so that there are no relativistic

e�ects), at speed u: During the time t between one emission and the next,

the source \catches up" a distance ut with its previous wavefront, so the

distance � between wavefronts is no longer c=� but instead

� =c

�� ut =

c

�� u

�

=c

�(1� �) :

Its apparent frequency is therefore

�0 =

c

�=

�

(1� �):

30

This is the classical Doppler shift (as it applies to sound, for example). But

when the source is moving towards us, its \clock" runs more slowly by a

factor = 1=p1� �2; so the rate at which it emits pulses is no longer � but

�= . Therefore, the frequency actually shifts to

�0 =

�

(1� �)(3.6)

= �

s1 + �

1� �:

This is the equation for the relativistic Doppler shift. A classic example

is the redshift seen due to the expansion of the universe. Hubble was the

�rst to observe that the further away a galaxy lies from us, the faster it is

moving away. As it moves away, the frequency of the light it emits drops,

and the wavelength therefore increases towards the red end of the spectrum.

(The opposite e�ect, the increase in frequency as sources approach, is called

\blueshift"). From the distances and speeds of the galaxies, and taking into

account the reduction in gravitational attraction as objects move apart, the

age of the universe has been calculated to be about 14� 2 billion years.

Note that (3.6) gives the measured frequency in a moving frame, starting

with the source frequency. Often, of course, the source will be in the moving

(primed) frame, and the shift is then inverted; just as with the Lorentz

contraction and time dilation, it is important to think and to make sure that

the shift is in the right direction. Later, we shall derive an expression for the

Doppler shift for light emitted at a general angle � to the direction of motion

of the source.

3.7 Synchronisation of Clocks

Einstein, at age 14, is reported to have wondered what it would \look like"

to ride along with a beam of light. He realised that time would appear to

be \frozen"; if the light was emitted from a clock which said 12:00 noon, the

image of that light would still say 12:00 noon when it arrived at the observer,

however far away he was... This raises the question of what the time \really"

is, and how we should measure it in a consistent way.

Events can be measured by a set of synchronised clocks and rigid rods

with which the reference frame is provided. There are two ways in which we

can synchronise distant clocks:

31

1. We can synchronise them when they are right next to each other and

then separate them very slowly so that the time dilation correction is

negligible.

2. We can send a beam of light from one clock to another. Suppose we

send a light beam out in this way from clock A, which reads time t1;

to clock B and back (see diagram). Clock B reads time t2 when the

light pulse arrives, and A reads t3 when the pulse returns to its starting

point. The clocks are synchronised if t2 is equal to the average of t1and t3, in other words

t2 =1

2(t1 + t3) :

BA

t2

t3

t1

Figure 3.3: How to synchronise remote clocks.

This just allows for the travel time of the light between the clocks.

32

3.8 Summary

We have seen how (a) lengths contract along the direction of motion, and (b)

moving clocks slow down. This is true for all observers in relative motion: just

as A sees B's clocks slow down, so B sees A's clocks slow down. The apparent

discrepancy is resolved because to measure the length of a moving object,

you have to measure simultaneously the positions of the two ends, which

are spatially separated; however, the observers disagree about the timing

of spatially-separated events, and the disagreement in time measurements

exactly compensates the disagreement in length measurements between the

two frames.

33

Chapter 4

Spacetime

4.1 Spacetime Events and World Lines

We often represent space and time on a single spacetime diagram, with time

on the vertical axis (see diagram). An event is something that happens at

a given point in space and time, and is represented by a point E (x; t) on a

spacetime diagram.

t

x

E(x,t)

Figure 4.1: Spacetime diagram of an event.

34

The path traced out by an object in a spacetime diagram is called its

world line.

t

x

AB

θ

Figure 4.2: Spacetime diagram showing world lines of stationary and moving

objects.

� A vertical line (A in diagram) represents a stationary object.

� An object moving with velocity v is represented by a line at angle � to

the horizontal (B in diagram), where

v = cot �:

� A light ray has cot � = c; we usually measure x in units of \ct" (e.g.

\light seconds"), so � = 45Æ for photons. Note that nothing can have

a world line with � < 45Æ.

4.2 Intervals

Consider the quantity

S2 = c

2t2 �

�x2 + y

2 + z2�: (4.1)

35

The last three terms are the distance (squared) from the origin to the

point at which an event occurs; the �rst term is the distance that light can

travel in the available time. Let us evaluate the same quantity in another

frame of reference, moving with velocity v and having the same origin when

t = t0 = 0:

S2 = c

2 2

�t0 +

vx0

c2

�2

� 2 (x0 + vt

0)2 � y

02 � z02

= 2

�c2t02

�1� v

2

c2

�+ 2vt0x0 � 2vt0x0 � x

02

�1� v

2

c2

��� y

02 � z02

= c2t02 �

�x02 + y

02 + z02�= S

02:

Therefore, the quantity S2, which is known as the interval, is the same in

all inertial reference frames; it is a Lorentz invariant. Remember this |

it's important, and very useful! The behaviour is very similar to the way

in which distance, in three-dimensional space, is invariant when we rotate

our coordinate axes; the interval stays unchanged when we move from x; t to

x0; t

0 axes via a Lorentz transformation.

Spacetime can be divided into three parts, depending upon the sign of

the interval:

1. S2> 0: Timelike (with respect to the origin). This class of events

contains x = 0; t 6= 0, which corresponds to changes in time of a clock

at the origin.

2. S2< 0: Spacelike (with respect to the origin). This class contains

t = 0; x 6= 0 events, which are simultaneous with but spatially separated

from the origin.

3. S2 = 0: Lightlike (with respect to the origin). Rays of light from the

origin can pass through these events. Consider a spherical wavefront

of light spreading out from the origin; at time t it has radius r = ct, so

it is de�ned by the equation

x2 + y

2 + z2 = r

2 = c2t2;

which of course de�nes an interval S2 = 0. Naturally, in any other

frame, light also travels at speed c, so its wavefront must be determined

by the same equation, but using primed coordinates.

36

We can also de�ne intervals between two events, instead of relating it to

the origin. In this case,

S2 = c

2 (t2 � t1)2 � (x2 � x1)

2+ (y2 � y1)

2+ (z2 � z1)

2:

4.3 Simultaneity

Let's look more closely at the Lorentz transformations 3.5. The �rst three

are just the same as the Galilean transformations, except that we have the

length contraction factor in the direction of motion x. The fourth equation

looks similar to the Galilean transformation t0 = t with the time dilation

factor | but we now have an additional, unexpected term vx=c2: What

does it mean?

If, in the spaceship, there are two events that are separated in space |

suppose they occur at positions x1, x2 | but at the same time t0, then

according to our observer on Earth, they occur at times

t01 = (t0 � �x1=c) ;

t02 = (t0 � �x2=c) :

In that case, the events as seen from the Earth are no longer simultaneous,

but are separated by a time

t02 � t

01 = (x1 � x2) �=c:

This is known as failure of simultaneity at a distance, and it lies at the

heart of most of the problems and \paradoxes" of relativity. The whole

idea of simultaneity just breaks down: events (separated in space) that are

simultaneous in one reference frame are not in any other.

As an example, suppose the spaceship pilot is standing in the middle of

his spaceship, and he emits a ash of light which reaches both ends at the

same time. As seen by the man on Earth, though, the rear of the spaceship

is moving up to meet the light, whereas the front of the spaceship is moving

away from it; so the backwards-going pulse of light reaches the rear of the

spaceship before the forward-going pulse reaches the front (see diagram).

Note that from the point of view of a second rocket which is overtaking

the �rst, the �rst one is moving backwards, in which case the light will hit

the front mirror �rst.

37

(b)

(a)

Figure 4.3: An example of the failure of simultaneity at a distance.

In order to appear simultaneous in all Lorentz frames, a pair of events

must coincide in both space and time | in which case they are really just

one event.

4.4 The Principle of Causality

Easily stated | this is simply that

causes always occur before their resultant e�ects.

If A caused B, then A happened before B. That's all. It prevents nasty

paradoxes: for example, you can't go backwards in time and prevent your

own birth, because if you did, you would never have been born, and so you

could not have travelled back... and so on. We don't have any proof of it;

it's just an observation of the way things seem to be. It keeps life simple.

But in relativity, it has an interesting consequence.

Look again at the rocket in the previous section. In its frame of reference,

light hits each mirror at the same time. In Earth's frame, light hits the back

mirror �rst. In the frame of an overtaking rocket, light hits the front mirror

�rst. Suppose that, just as light hits the back mirror, a guard standing

beside it drops his sandwich; and just as light hits the front mirror, the pilot,

38

standing beside it, spills his co�ee. Is it possible that the guard dropping his

sandwich caused the pilot to spill his co�ee? Although it might seem like

it from the Earth, it clearly cannot be, since from the overtaking rocket the

pilot seemed to spill his co�ee �rst. The order of the sequence depends upon

the frame of reference, and so there is no cause-and-e�ect relationship.

On the other hand, if one event does cause another, it will occur �rst in all

reference frames.This can happen if and only if the interval between them

is timelike or lightlike; in other words, for one event to in uence another,

there has to have been enough time for light to propagate between them.

Therefore, it is clear that nothing | no information | can travel faster

than the speed of light. In fact,

� If events are physically related, their order is determined absolutely.

� If they are not physically related, the order of their occurrence depends

upon the reference frame.

(N.b. \physically related" here means that information has had time to

travel between the events, not necessarily that one caused another).

4.4.1 Example: Gun�ght on a Train

Another example. Imagine a train that moves at 0.6 c: A man on the ground

outside sees man A, at the rear of the carriage, start shooting at B, who is

standing about 10 m ahead of him. After 12.5 ns, he sees B start shooting

back.

But the passengers all claim that B shot �rst, and that A retaliated after

10 ns! Who did, in fact, shoot �rst?

The answer depends upon the frame of reference. Since the light could

not travel the 10 m between the protagonists in the 10 ns or so required,

the events are spacelike | there is no cause-and-e�ect relationship, and the

sequence is relative.

4.4.2 Example: A Lighthouse

Suppose two satellites are positioned 6� 108 m apart, and midway between

them is a lighthouse, which rotates once every two seconds. The satellites are

programmed to start broadcasting when the beam of the lighthouse sweeps

39

past them. The \spot" of light passes one satellite, which duly begins trans-

mitting; one second later, the spot of light has travelled halfway around the

circle of radius 3 � 108 m, and it reaches the second satellite, which also

begins broadcasting. The light spot has travelled 109 m in just one second

| a speed of 10c; and the satellites also began transmitting within a second

of each other, despite being separated by 6 � 108 m | an apparent trans-

mission of information at 2c. Is this a problem? No: in fact, no information

has passed between the satellites. They are merely responding to informa-

tion that has been transmitted from the central point at speed c. If the �rst

satellite's antenna had failed and it had not begun broadcasting, the second

satellite would have known nothing about it, and would have started its own

transmission regardless. The two events are not physically related.

4.5 Light Cones

If we include another space dimension in our spacetime diagram, the \world

line" of light then de�nes a cone (actually a pair of cones), with its vertex at

the origin and an opening angle of 45Æ.

Events at the origin can be related to events inside the cone, since signals

can travel between them at up to the speed of light; but an event at the

origin cannot be related to events outside the light cone.

The region inside the cone and \above" the origin is therefore known as

the absolute future, and that inside the cone below the origin is the absolute

past.

4.6 Visual Appearance of Moving Objects

Consider a cube moving past, close to the speed of light. Naturally, it is

Lorentz contracted, and this can be measured, for example, by the set of

clocks (at rest and synchronised in the laboratory frame) which all read the

same time at the moment that the corners of the cube pass them. In this

way, the time lags for light to travel from the di�erent corners of the cube

are eliminated.

But our eye does not work like this! It can only be in one place at a time,

and it registers only light that enters at the same time. Hence, what one sees

may be di�erent from the measurements made by a lattice of clocks.

40

When it is at 90Æ (see �gure), photons from the nearest corners of the cube

arrive simultaneously, and so one will see the normal Lorentz contraction of

the bottom edge. (Here we assume that the cube is quite far away). However,

light from one of the further rear corners (E in the diagram) that left that

point earlier can arrive at the eye at the same time! The observer sees behind

the cube at the same time | and it therefore appears to him to be rotated.

Figure (b) shows the appearance of the cube as the observer looks up at

it; Figure (c) shows how the observer might interpret it as a rotation.

Di�erent con�gurations can be far more complicated. A rod that is ap-

proaching an observer almost head on will appear to be longer, despite the

Lorentz contraction, because light emitted from the rear takes some time to

\catch up with" light emitted from the front.

A diagram is included to show the approximate appearance of a plane

grid, moving at relativistic speeds past an observer. The observer is at unit

distance in front of (\above") the origin, i.e. the line from observer to origin

is perpendicular to the direction of motion. (This diagram is a rough copy

of a plot from Scott, G.D., and Viner, M.R., Am. J. Phys. 33, 534, 1964).

4.7 Oblique Axes

Consider a clock A moving in reference frame S. Its world line is at angle

�0 = cot�1 v to the horizontal (see �gure 4.7).

In frame S 0 moving with the clock, the clock is at a �xed x0, so its world

line must be parallel to the time axis t0.

The x0 axis is determined by

t0 = 0

) t =�x

c;

and therefore it is at an angle �x0 = tan�1 (�=c) to the x axis.

Axes x0; t0 are oblique in this representation. The coordinates of an event

are determined by drawing lines parallel to the axes (see diagram). This

procedure reduces to dropping perpendiculars if the axes are orthogonal.

Oblique axes are essential to represent the coordinates of one event in

di�erent frames... but one cannot think of \distances" between points as

that obtained by direct measurement with a ruler!

41

As an example, let's look again at the issue of simultaneity. We considered

a man at the centre of a spaceship (B) emitting a ash of light to the rear

(A) and the front (C) of his spaceship. In his frame of reference, the world

lines are as shown in the �rst diagram, with the light reaching the two ends

of the spacecraft simultaneously.

From the Earth's point of view, with the spaceship moving past, the world

lines are tilted. Light, however, still travels along its 45Æ world lines. We can

see that the light will intercept the rear of the spaceship before it reaches

the front. Clearly, in the spaceship's frame of reference, the rear of the craft

is not moving, so the time axis t0 is parallel to A's world line; and, since in

the rocket the light reaches both ends simultaneously, the axis of constant

time (x0) must be parallel to the line joining the two events where the light

crosses the world lines A and C.

This representation provides useful imagery, but it is not very practical

for solving problems! You won't need it for your exams.

4.8 More Paradoxes

4.8.1 The Pole and Barn Paradox

A pole vaulter carries a 20-metre long pole. He runs so incredibly fast that

it is Lorentz contracted to just 10 m in Earth's reference frame. He runs into

a 10 m long barn; just at the moment when the pole is entirely contained

inside the barn, the doors are slammed closed, trapping both runner and pole

inside.

However, from the runner's point of view, it is the barn that is contracted

to half its length. How can a 20-metre pole �t inside a barn that appears to

be just 5 m long?

The answer, of course, lies in the failure of simultaneity. In the barn's

reference frame the doors are closed at the same instant; to the poor runner,

however, it appears that the exit door is closed, and his pole runs into it,

before the rear of his pole has completely entered the barn.

4.8.2 The Thin Man and the Grid

A man (perhaps our speedy pole vaulter) is running so fast that Lorentz

contraction makes him very thin. Ahead of him in the street there is a grid.

42

A man standing beside the grid expects the thin runner to fall through one

of the spaces in the grid. But to the runner, it is the grid that is contracted,

and since the holes are much narrower, he does not expect to fall through

them. What actually happens?

This is actually a rather subtle problem that hinges on the question of

rigidity. There is, in fact, no such thing as a perfectly rigid rod. Consider

a bridge from which the support at one end is suddenly removed. That end

starts to fall at once. The rest of the bridge, however, stays as solid as ever,

until the information that the support is missing reaches it | in this case,

with a speed determined by the time required for an elastic wave to move

through the steel.

Consider next a rod lying on a ledge in a rocket. The ledge suddenly

collapses and the rod falls down with the acceleration of gravity. But in the

Earth's frame, the end at the back of the rocket starts falling �rst, before

the ledge at the front end has collapsed at all. The rod thus appears bent |

and is bent | in Earth's frame.

As for the man and the grid, let us replace the problem with a \rigid"

metre-long rod sliding along a table in which there is a metre-wide hole. In

the frame of reference of the hole (grid), the rod (man's foot) simply falls into

the hole. From the point of view of the rod (runner), though, the front end of

the rod (foot) droops over the edge of the hole when the support underneath

vanishes, and the rest of it comes following after.

4.8.3 Twins Paradox Revisited

Here we look again at the twins paradox, using a slightly di�erent approach.

We will not have time to go through this in the lectures; it may take a little

while to understand what is happening to the clocks in each reference frame,

but it is worth the e�ort.

Twin B leaves twin A with relative velocity v, reverses his velocity and

returns to �nd less time has elapsed on his clock than on A's. To A, this is

in agreement with time dilation, but B saw A moving with respect to him

and so thinks that A's clock should show less time elapsed.

To avoid e�ects of accelaration, we introduce a third \twin" C who has

velocity �v with respect to A and who coordinates his clock with B as they

cross. We now list the (x; t) coordinates of the important events in the various

inertial frames:

43

A's frame B's frame C's frame

B leaves A 0; 0 0; 0 �2vt= ; 0B and C cross vt; t 0; t= 0; t=

C returns to A 0; 2t �2vt= ; 2t= 0; 2t=g

We see that the total time elapsed in frame A is 2t, compared to the time

elapsed in B and C which is 2t= . Therefore, a greater time has elapsed in

A than in B or C, as expected.

The paradox can be resolved by considering the time on A's clock at the

instant when B and C meet. There are two events to consider: (1) B meeting

C, (2) recording the time on A's clock. These events are spatially separated,

and therefore subject to the usual failure of simultaneity at a distance. The

crucial point is that since A is spatially separated from the B, C meeting

point, the reading on A's clock will depend upon the frame of reference in

which the simultaneity between the two events is assumed. Thus:

A's frame: A's time is clearly t, coincident with the point (vt; t) in A's

frame at which B and C meet.

B's frame: In this frame the appropriate coordinates of A at the meeting

time are (�vt= ; t= ), which transform to (0; t= 2) in A's frame; in other

words, from B's point of view, the B-C crossing occurs simultaneously with

A's clock reading t= 2. Since t= 2 < t= , from B's point of view A's clock

is going more slowly, as expected.

C's frame: In this frame the appropriate coordinates ofA are also (�vt= ; t= ),which in this case are transformed to (0; 2t� t= 2) in A's frame. Again the

change in A's clock during the return journey is t= 2 < t= , and therefore

from C's point of view also, A's clock is going more slowly.

In other words, both B and C think that A's clock is going more slowly;

and they do set their clocks to agree with each other at the point at which

they cross; but because A's clock is spatially separated from their meeting

point, the relative velocity between B's and C's frames gives rise to a dis-

agreement about the reading on A's clock that is simultaneous with their

meeting.

A's various clock readings are:0 Start

t= 2 B and C meet, according to B

t B and C meet, according to A

2t� t= 2 B and C meet, according to C

2t Finish

The discrepancy 2t � 2t= 2 between B and C is of just the right value

44

to resolve the con ict between the view held by B and C that A's clock is

going more slowly and the actual result that C's clock reads less than A's at

the end.

45

xO

t

AbsolutePast

r = ct

AbsoluteFuture

45°

Figure 4.4: A light cone.

46

(1-β2)1/2

HE G

G

E

ϕH

E

HG

F

1

(b)

(c)(a)

β

Figure 4.5: (a) A cube passing by an observer, as seen in the laboratory

frame. (b) What the observer sees as he looks up. (c) How the observer

interprets what he sees.

47

Direction of motion

(b) v/c = 0.995

0

0

(a) v/c = 0

Figure 4.6: Distortion of a grid moving at relativistic velocities.

48

Tick... Event E(x,t) or E(x′,t′)

Clock A; vel. vt′

t

x

x′

θx′θ0

Tick... Event E(x,t) or E(x′,t′)

t′

(b)

(a)

x

x′

Figure 4.7: The oblique axes required to describe moving objects.

49

Chapter 5

Dynamics and Kinematics

5.1 Relative Velocities

A train moves at 0.8 c. A passenger �res a bullet which moves at 0.6 c relative

to the train. How fast does it move relative to the ground?

A Galilean transformation would give

v = 0:8c+ 0:6c = 1:4c:

This is clearly wrong...

5.2 Lorentz Transformation of Velocities

Consider an object moving with velocity u = uxbx + uyby in reference frame

S. (Note that the transverse directions by and bz are equivalent). So,ux =

dx

dt; uy =

dy

dt:

In frame S 0, moving with velocity v along the x axis, the same object will

have velocity

u0x=

dx0

dt0; u

0y=

dy0

dt0:

Since

x = (x0 + �:ct0)

y = y0

ct = (ct0 + �x0) ;

50

we have

dx = (dx0 + �:cdt0) = (u0

x+ v) dt0

dy = dy0 = u

0y:dt

0

dt = �dt

0 + v:dx0=c

2�=

�1 + v:u

0x=c

2�dt

0:

Therefore,

ux =dx

dt=

u0x+ v

1 + v:u0x=c2

(5.1)

uy =dy

dt=

u0y

(1 + v:u0x=c2)

uz =dz

dt=

u0z

(1 + v:u0x=c2)

and, correspondingly,

u0x

=ux � v

1� v:ux=c2

(5.2)

u0y

=uy

(1� v:ux=c2)

u0z

=uz

(1� v:ux=c2):

Note that velocities in the transverse direction are altered.

We can now answer the question about the bullet in the train; an object

moving with velocity u0x= 0:6c in a frame that is itself moving at velocity

v = 0:8c moves (in our, unprimed, frame) at a velocity

ux =u0x+ v

1 + v:u0x=c2

=(0:6 + 0:8) c

1 + (0:6) (0:8)= 0:95c:

What about light itself? Suppose, instead of a bullet, the man �res a laser;

the photons move at u0x= c: In this case, we �nd that

ux =(1:0 + 0:8) c

1 + (1:0) (0:8)= c;

as we expect. This is a nice cross-check.

51

5.3 Successive Lorentz Transformations

Consider three reference frames, S; S 0 and S00 | say, of someone standing

on the ground; of someone moving past in a train; and of someone moving

past in a train. What happens if we Lorentz transform from the �rst to the

second frame, and then again from the second to the third? Is it consistent

with transforming directly from the �rst to the third frame?

Let S 0 move along the x axis of frame S with velocity v = �c; so

x0 = (x� �ct)

ct0 = (ct� �x) :

Likewise, let S 00 move along the x0 axis (which is the same as the x axis) with

velocity v0 = �

0c, so

x00 =

0 (x0 � �0ct0)

ct00 =

0 (ct0 � �0x0) :

We can substitute for x0 and t0 to give

x00 =

0 ( (x� �ct)� �0 (ct� �x))

= 0 (1 + ��

0) x� 0 (� + �

0) ct;

ct00 =

0 ( (ct� �x)� �0 (x� �ct))

= 0 (1 + ��

0) ct� 0 (� + �

0) x:

Since, in S; the frame S 00 is moving with velocity

v00 =

v + v0

1 + v:v0=c2;

it can be shown (with a few lines of messy algebra) that the corresponding

factor is

00 =

0 (1 + ��0) :

Therefore, we �nd that

x00 =

00 (x� �00ct)

ct00 =

00 (ct� �00x) :

In other words, two successive Lorentz transformations give another Lorentz

transformation, corresponding to the appropriate relative velocity.

52

5.4 Velocity Parameter

You will sometimes come across a quantity known as the velocity parameter.

This is analagous to the idea of adding angles instead of slopes in normal

geometry; two angles �1; �2 correspond to slopes S1 = tan �1; S2 = tan �2;

but to add the angles we use

�tot = �1 + �2

whereas the resulting slope has a more complicated addition law:

Stot = tan �tot =tan �1 + tan �2

1� tan �1 tan �2:

This is very similar to our formula for the addition of velocities, and indeed,

we �nd that if we de�ne a velocity parameter � such that

� = tanh �

then the parameter � adds linearly like an angle. Notice that we have to use

the hyperbolic tangent tanh, rather than an ordinary \tan", because of the

same signs in numberator and denominator in 5.1. Tanh is de�ned as

tanh � =e� � e

��

e� + e��:

In practice, we will not make much use of this function; you won't need it

for your exams.

5.5 Relativistic Dynamics

As we pointed out at the beginning, our classical laws of conservation of

momentum and energy require modi�cation in the relativistic limit. However,

we can obtain laws closely related to the originals if we allow \mass" to vary

with velocity in the way that we have allowed lengths and times to vary.

Let us look at a \bomb" (or particle) of massM0, which breaks up into two

fragments, each of mass mu that move o� with equal and opposite velocities

u (see diagram).

53

(a) From rest frame S’ of initial object...

mu mu

u u

M0

(b) From rest frame S of left-going fragment.

mu mv

u v

Mu

Figure 5.1: The breakup of a massive object, seen from the reference frames

of (a) the parent object, (b) one of the fragments.

Now, let us move to the rest frame of the left-going fragment (see dia-

gram). In that frame, the initial particle is moving to the right with velocity

u, whereas the other fragment is moving even faster to the right, with velocity

v =u+ u

1 + u:u=c2=

2u

1 + u2=c2;

according to our law of addition of velocities. Since we are allowing masses to

vary with velocity, we now call the initial mass Mu; the stationary fragment

has mass m0 and the other fragment has mass mv:

We assume that

1. something that we shall call the total \mass" is conserved;

2. \momentum" (= mass x velocity) is also conserved.

Therefore,

mv +m0 = Mu

mv:v + 0 = Mu:u

so

mv +m0 = Mu = mv:v=u

= mv:2

1 + u2=c2:

54

Therefore,

m0 = mv

�2

1 + u2=c2� 1

�= mv

�2� 1� u

2=c

2

1 + u2=c2

�(5.3)

= mv:1� u

2=c

2

1 + u2=c2:

But we need to know it in terms of v, not u. Let us consider our old friend

1� v2=c

2 = 1� 4u2=c2

(1 + u2=c2)2

=1 + 2u2=c2 + u

4=c

4 � 4u2=c2

(1 + u2=c2)2

=(1� u

2=c

2)2

(1 + u2=c2)2:

But this is just the square of the ratio of \stationary" to \moving" masses

in 5.3. So, we �nally obtain

mv =m0p

1� v2=c2= m0

as the quantity that we called \mass" and that is conserved in this reaction.

It acts just as though mass itself were increasing with velocity { and indeed, in

many textbooks \relativistic mass" is de�ned this way, with m0 being called

the \rest mass". However, for consistency, it is best (and more common

nowadays) to consider the mass to be de�ned as being measured at rest, and

the is always written explicitly when required. From now on, m or m0

will refer to the rest mass only. We shall drop the notation mv and

the idea of \mass" varying with velocity; so-called \relativistic mass" will be

written explicitly as m.

For the above problem, then, if we rede�ne the (rest) mass of the parent

to be M , and the (rest) mass of each of the daughter products to be m, we

�nd that everything is consistent if M = 2 m:

What happens to this quantity m at low velocities? Let us do a binomial

expansion:

m = m�1� v

2=c

2��1=2

= m+1

2m:v

2:1

c2+3

8mv4

c4:::

55

The second term is the classical kinetic energy, divided by the constant c2:

This suggests that, if we multiply the entire expression by c2; we will obtain

an energy:

E = mc2 = mc

2 +1

2mv

2 +3

8mv4

c2::: (5.4)

(We have dropped the subscript v from m). The �rst term of this expan-

sion is called the \rest energy", and the second term is, as we noticed, the

classical kinetic energy. However, we see that the true kinetic energy { liter-

ally, movement energy, so total energy minus the energy that the object has

anyway when it's standing still { is

T = E �mc2

=1

2mv

2 +3

8mv4

c2+ :::;

and it actually has higher-order terms contributing; the classical expression

is just a �rst approximation.

If we de�ne a momentum

p = mv;

we see that

p2 +mc

2 =m

2v2

(1� �2)+m

2c2:(1� �

2)

(1� �2)

=m

2v2 +m

2c2 �m

2v2

(1� �2)

=m

2c2

(1� �2)

= 2m

2c2 = m

2c2 = E

2=c

2;

so

E2 = p

2c2 +m

2c4: (5.5)

This equation is extremely useful in relativity! The quantity E2�p2c2 = m2c4

just depends on the total rest energy of an object (or system), and so is

independent of the reference frame in which it is measured, in just the same

way as the interval S2 that we met earlier.

Another equation that is sometimes useful:

p:c = mv:c = E:�

56

Finally, note that if a particle is in a potential of some sort, its potential

energy of course must add to the expression 5.4 for the total energy.

5.6 Consequences

Equations 5.4 and 5.5 have a lot of interesting consequences...

� Equivalence of mass and energy. We postulated a \conservation of

mass" law, but it has turned out that mass and energy are the same

thing, just measured in di�erent units (and related by the constant c2).

So the law has become equivalent to the conservation of energy.

� Notice that the rest mass is not conserved. When the object of mass

M0 split up, it produced two objects each of so-called\mass" (actually

times mass) mu; but the sum of the rest masses of the two frag-

ments is 2m0, which is less than M0; the \extra" mass in M0 has been

\converted" into kinetic energy of the fragments. The same is true in

reverse; if we put two fragments of mass m0 together gently, we end

up with an object of mass 2m0 | which is therefore a di�erent object

than we get if we push the two fragments together hard, since the extra

kinetic energy \turns into" extra mass.

� If an object splits up in this way, when the fragments interact with

material, they deposit energy (as heat, kinetic energy and so on) until

they come to rest. The total energy \liberated" then is

M0c2 � 2m0c

2:

The atomic bomb worked so well because the fragments (iodine, xenon

etc) produced when the uranium atom splits are so much lighter than

the initial uranium atom itself.

� The principle also works \in reverse": if you want to split up a carbon

dioxide molecule into carbon and oxygen atoms, you have to add energy

| so the sum of the masses of the carbon and two oxygen atoms is

actually greater than the mass of the carbon dioxide molecule. (The

sun is powered by the energy liberated when hydrogen atoms \fuse"

together to make helium: it is losing mass at a rate of about 5 million

tons per second).

57

� We see that a little mass is equivalent to a lot of energy. One gram of

material is equivalent to 9� 1013 J of energy | that's about 3 MW of

power for a year, or the equivalent of 20 ktons of TNT.

� A peculiar consequence, without any particular practical use: mass

can be transferred without exchange of either particles or radiation.

Consider a board, on one end of which is a motor. This turns a drive

belt, which itself turns a rotor at the other end of the board. Energy,

and therefore mass, is transferred from the motor to the rotor, i.e., from

one end of the board to the other. If it were mounted on frictionless

rails, it would accelerate in the opposite direction so as to conserve

momentum.

� What about light? Since

E = mc2;

but !1 as v ! c, the only way we can stop this from diverging is

if photons have \zero rest mass": m = 0. They can still carry energy,

though, by virtue of their momentum; from 5.5, we see that for photons,

E = pc:

What happens if you \stop" a photon? You cannot! A photon always

travels at the speed of light!

5.7 Lorentz Transformation of Energy-Momentum

Consider a dynamic \event" like the particle breakup we saw earlier. What

do the energy and momentum look like from a di�erent coordinate system? It

turns out, in fact, that they transform in a manner very similar to space and

time. Here we will derive the transformation in two di�erent ways; �rstly,

using familiar ideas but quite a lot of tedious algebra; and then in a much

simpler way, treading on perhaps slightly less familiar territory. Let us �rst

map out the road we shall follow.

5.7.1 The DiÆcult Way

This way is rather heavy in algebra, and I won't go through it in the lectures.

However all of the elements are familiar, so there aren't any tricky concepts

to get hold of.

58

� Starting point: Imagine a spaceship moving (relative to the Earth) with

velocity v along the x axis, and a particle moving with velocity u.

� Notation: refers to the spaceship, not the particle; i.e. it is (1� v2=c

2)�1=2

not (1� u2=c

2)�1=2

: The quantities u0; E 0, 0; p0 are the velocity, energy,

\gamma" and momentum of the particle as seen from the spaceship.

Earth's reference frame is S, the spaceship's is S 0:

� Goal: to calculate E 0; p

0 in terms of E; p:

� Procedure:

{ We begin with the x component; let u = ux:

{ Find u0 and then

0 in terms of u and v and then substitute into

E0 =

0m0c

2

p0x

= 0m0u

0:

{ For the transverse components, we consider a particle moving in

the y direction, and �nd a new 0 based on the new u

0. From this,

we can calculate p0y(and check that the energy transformation is

still valid).

Okay, here we go...

The velocity u0 of the particle as seen from the spaceship is (from eqn

5.2)

u0 =

u� v

1� uv=c2:

Now let us calculate the energy E 0 of the particle as seen from the spaceship.

The rest mass, of course, is the same; but we need to recalculate in order

to �nd the new mass. We start with

u02 =

u2 � 2vu+ v

2

1� 2vu=c2 + v2u2=c4

so

1� u02=c

2 =(1� 2vu=c2 + v

2u2=c

4)� (u2=c2 � 2vu=c2 + v2=c

2)

(1� vu=c2)2

59

=1� u

2=c

2 � v2=c

2 + v2u2=c

4

(1� vu=c2)2

=(1� u

2=c

2) (1� v2=c

2)

(1� vu=c2)2

and therefore

0 =

1p1� u02=c2

=1� vu=c

2p(1� u2=c2)

p(1� v2=c2)

:

The energy is

E0 =

0mc

2

=mc

2 �muvp(1� u2=c2)

p(1� v2=c2)

:

Since E = mc2=p1� u2=c2, and the original = 1=

p1� v2=c2; we obtain

E0 = E � v

mup(1� u2=c2)

(5.6)

= (E � vpx) :

This looks familiar.... Let us now consider the momentum, �rstly in the x

direction.

p0x

= 0m

0u0 =

E0

c2u0

=1

c2� mc

2 �muvp(1� u2=c2)

p(1� v2=c2) � u� v

1� vu=c2

=mp

(1� u2=c2)� (1� uv=c

2)p(1� v2=c2)

� u� v

(1� vu=c2)

=mu�mvp

(1� v2=c2)p(1� u2=c2)

= �px � v:E=c

2�: (5.7)

What about transverse components? Consider a particle moving upwards

with velocity u in frame S: If we transform to frame S 0, moving with velocity

v in the x direction, we �nd the new velocity components are

u0x

= �vu0y

=uy

60

where, as usual, = (1� v2=c2)�1=2 (not to be confused with (1� u2=c

2)�1=2

| we aren't transforming to the rest frame of the particle). So, the total

velocity squared is

u02 = v

2 +u2y

2;

and

1� u02=c

2 = 1� v2=c

2 + u2

y=c

2(1� v2=c

2)

= (1� v2=c

2)(1� u2=c

2):

This gives

0 =

1p(1� v2=c2)

p(1� u2=c2)

:

as before. Therefore, the transverse component of momentum is

p0y

= 0mu

0y

=mp

(1� u2=c2)

1p(1� v2=c2)

uy

= m:uy

In other words, the transverse component of momentum is unchanged. Let's

look at the energy again for this case:

E0 =

0mc

2

=mc

2p(1� u2=c2)

1p(1� v2=c2)

= E:

This time, there is no vp term, so we were correct previously in writing the

transformation using the x component only.

In total, then, the energy and momentum transformations are

p0x

=

�px � �:

E

c

�(5.8)

p0y

= py

p0z

= pz

E

c

0

=

�E

c� �px

�:

61

It is no coincidence that this set of transformations is just like those for

x and t. If we replace E=c by ct and px by x; etc., we regain the original

transformations.

Recall that the interval

S2 = c

2t2 � x

2 � y2 � z

2

was an invariant under Lorentz transformations. If we make the same sub-

stitutions here, and then divide by c2; we obtain

E2 � p

2c2 = const = m2

0c4:

The invariant corresponding to the interval is therefore the rest energy of the

object, which is naturally the same whichever frame you calculate it in.

5.7.2 The Easy Way

Now we shall derive the same transformation, but rather more quickly and

easily. This is the way it will be done in the lecture. Recall that the interval

S2 = c

2t2 �

�x2 + y

2 + z2�

is invariant under Lorentz transformations | it's the same in all inertial

reference frames. Now, for a particle travelling a short distance u�t in time

�t,

�x2 +�y2 +�z2 = u2�t2;

so the interval between the start and the end of the journey is

�S2 = c2�t2 � u

2�t2

= c2�1� �

2��t2

= c2 � �t

2

2u

: (5.9)

(The time

� =�t

u

is the proper time | the time as measured in the particle's own rest frame,

and so it is quite natural that the quantity (5.9) is the same from whatever

frame it is calculated).

62

Knowing that �t= and m0 are both invariant, we construct

k = um0

�t= k

0;

which must also be invariant under the Lorentz transformation | i.e., it's

the same quantity in all frames: k = k0.

Let us go back to our particle, and see how its short journey looks from

another reference frame. Under the usual transformation, we �nd

�x0 = (�x� �:c�t)

�y0 = �y

c�t0 = (c�t� �:�x)

where, of course, � and refer to the velocity of the new reference frame

rather than the particle.

Multiplying throughout by the Lorentz-invariant quantity k (or k0, which

is equivalent), we �nd

k0�x0 = (k�x� �:ck�t) (5.10)

k0�y0 = k�y

ck0�t0 = (ck�t� �:k�x) :

But

k�x = um0

�x

�t= um0ux = px;

and, likewise, k�y = py; k0�x0 = p

0xetc. Also,

ck�t = um0 =E

c2

ck0�t0 =

E0

c2:

So, from (5.10), the Lorentz transformations for energy and momentum are

p0x

=

�px � �:

E

c

�p0y

= py

p0z

= pz

E

c

0

=

�E

c� �px

�(5.11)

63

as we found above. Again, it is easy to show that

E2 � p

2c2;

which is simply the rest mass squared times c2, is invariant under the Lorentz

transformation.

5.7.3 Matrix Notation

Using matrices, the transformation (5.11) may be written as0BB@p0x

p0y

p0z

E

c

0

1CCA =

0BB@ 0 0 �� 0 1 0 0

0 0 1 0

�� 0 0

1CCA0BB@

px

py

pzE

c

1CCA :

5.8 Example: Collision Threshold Energies

Consider the reaction

p+ p! p+ n+ �+;

in which an incoming proton p (mass � 938 MeV/c2) of (total) energy Ep

hits a target proton at rest, to create a proton, a neutron (also of mass �938 MeV=c2) and a positive pion (mass 139 MeV=c2). What is the minimum

(threshold) kinetic energy that the incident proton requires for this reaction

to take place?

� The Trap to Avoid: the simple, obvious answer | that the sum of

the masses afterwards is 139 MeV greater than the sum of the masses

before, so this is the energy required | is wrong. This is because the

incident proton carries some momentum, and so the outgoing particles

must carry some momentum too.

� The Way to Solve it: A standard trick for many relativity \colli-

sion" problems | see, for example, our earlier \derivation" of the

energy-momentum transformations | is to transform into the centre-

of-momentum (or zero momentum) frame (note, this is no longer nec-

essarily the \centre of mass"!). In this frame the two protons rush

together with equal and opposite momenta, carrying just enough en-

ergy to make the three �nal particles (but not to give any of them

64

any kinetic energy). Now, you can use conservation of energy and mo-

mentum, solve the resulting equations, and transform back to the lab

frame; but the easy way is to use the fact that E2�p2c2 is invariant: itis the same in the laboratory and in the centre-of-momentum system.

In the lab frame, looking at the system before the collision, we have

Totalenergy = Etot = Ep +mpc2

Totalmomentum = pp =

�E

2p

c2�m

2

pc2

�1=2

:

In the centre-of-momentum system, the two protons come together with equal

and opposite momenta, to give a total of zero momentum; and, as we are at

threshold, the particles that are created in the collision are created at rest.

Therefore,

Totalenergy = E0tot

= mpc2 +mnc

2 +m�c2

Totalmomentum = 0:

Now we simply equate E2 � p2c2 in the two frames:

�Ep +mpc

2�2 � �E2

p

c2�m

2

pc2

�c2 =

�mpc

2 +mnc2 +m�c

2�2 � 0:

Expanding the brackets, we obtain

E2

p+ 2Epmpc

2 +m2

pc4 � E

2

p+m

2

pc4 = (mp +mn +m�)

2c4

2Epmpc2 + 2m2

pc4 = (mp +mn +m�)

2c4

Ep =(mp +mn +m�)

2c2

2mp

�mpc2:

Thus, the total energy of the incoming proton is 1226 MeV; this includes

938 MeV \locked up" in the rest mass energy m0c2, and, therefore, a kinetic

energy of 288 MeV.

5.9 Kinematics: Hints for Problem Solving

At this stage, it's worthwhile summarising some useful formulae.

65

1. Conservation of Mass-Energy. The total mass-energy is always

conserved. Add up E = mc2 for all of the particles, and it is the same

before as after the collision.

2. Conservation of Momentum. Momentum is also conserved; but

remember, this is p = mv; the momentum of each particle is the

usual classical mv times the relativistic factor .

3. Invariance of E2 � p

2c2: Do you want to change reference frames

at all? For instance, to �nd out what's going on in the c.m. frame?

Remember that E2�p2c2 is the same in all inertial frames, and it appliesto the whole system as well as to individual particles. Furthermore, in

the c.m. frame, the total momentum is zero (by de�nition). Very, very

useful!

These three together will do most of your dirty work for you. To go

further, e.g. to �nd the easiest ways out when particles leave the collision at

various angles, we need four-vectors; but they come later.

5.10 Doppler Shift Revisited

From equation 5.8 we can see that, if we receive a photon of energy E = h�

that was emitted by a source moving with velocity u = �c in the x direction,

the energy of the photon in the source's frame of reference is

E0 = (E � �cpx)

= (E � �:E: cos �)

= E (1� � cos �) :

If the source is coming directly towards us, � = 0Æ, so we obtain

E0 = E (1� �)

= E(1� �)p1� �2

= E

s1� �

1 + �;

66

which is consistent with the expression we obtained for the Doppler shift

earlier, where we only considered this limited case. Be careful! The result

we obtained then referred to the received frequency in terms of the source

frequency, not the other way around, and so the primed and unprimed frames

are swapped in this discussion.

5.11 Aberration of Light Revisited

Consider a telescope on a \stationary" Earth, pointing at an angle � to

capture light from a star (see diagram). Now let the Earth move with velocity

v in the x direction. As before, we will have to tilt the telescope to a new

angle, �0, so that photons can travel down the telescope tube to the bottom.

v

(b)(a)

θ

Figure 5.2: Aberration of starlight.

Now,

tan � =py

px; tan �0 =

p0y

p0x

:

67

Using equations 5.8, we see that

tan �0 =p0y

p0x

=py

(px � v:E=c2): (5.12)

Since these are photons (and, as we have drawn it, moving in the �x; �ydirection),

px = �E=c: cos �py = �E=c: sin �:

Therefore,

tan �0 =sin �

(cos � + �)

Equation 5.12 is the relativistic formula for the aberration of light.

5.12 Natural Units

Before continuing, let us introduce natural units, whereby

c = ~ = 1:

This just means that, instead of using the arti�cial unit of a metre for distance

(originally de�ned as a ten-millionth of the distance from the pole to the

equator; now the distance that light travels in 1/299792458 s), we use the

\light-second"; c, if you like, is then one light-second-per-second. This seems

reasonable to do, as we have seen how space and time are intertwined. The

velocity u now becomes equivalent to �, and the Lorentz transformations

become

x0 = (x� ut)

y0 = y; z

0 = z

t0 = (t� ux) ;

where, as usual, = (1� u2)�1=2

: The interval is now

S2 = t

2 � x2 � y

2 � z2:

It is easier to write the equations this way, and easier to work with them |

and very simple to go back, if we have to, to equations with c all over the

68

place. Just look at the dimensions: for example, with the expression 1� u2,

we know that we cannot subtract a velocity squared, which has units, from

the pure number 1, so it is clear that we have to divide u2 by c2 in order to

make it unitless.

The same trick works with energy and mass, which are also obviously

di�erent aspects of the same thing. The equations for energy and momentum

are therefore

E = m = m0

p = mv = m0v;

and they transform as

p0x

= (px � uE)

p0y

= py; p0z= pz

E0 = (E � upx) :

The similarity between the space-time and the mass-energy Lorentz trans-

formations is now even more striking. We shall explore the similarity further

in the next chapter | although c will creep back in!

69

Chapter 6

Four-Vectors

We are used to the idea of vectors in three-dimensional space; quantities

like momentum, which we write p (in bold) to remind us that it has three

components; we are free to rotate or translate the axes as we like, but, even

though the individual components of the vector are all mixed up amongst

each other, the magnitude of the momentum stays the same. For example,

if we rotate by an angle �; the new coordinate system becomes

x0 = x cos � + y sin �

y0 = y cos � � x sin �;

and we see how the new x0 is a mixture of the old x and y. However, the

length squared l2 = x

2 + y2 is invariant. This is very similar to the way in

which Lorentz transformations treat space and time.

Is there, therefore, some way in which we can put the separate components

of space and time together to come up with a function that is invariant under

Lorentz transformations? There is, in fact; it is called a four-vector, written

A�, where � indicates any of the four components. For spacetime,

x1 = x

x2 = y

x3 = z

x4 = ct;

and the transformation is

x01 = x1 � �x4

70

x02 = x2; x

03 = x3

x04 = x4 � �x1:

The interval is then

S2 = x

2

4 ��x2

1 + x2

2 + x2

3

�:

There are di�erent conventions; sometimes, the time component is called

x0 instead of x4, and sometimes this component is written with an i:

x4 = ict;

just so that the components will add in quadrature like normal vectors (so,

e.g., the interval S is given by �S2 = x21 + x

22 + x

23 + x

24). We shall not be

using that convention, but you should be aware that it exists.

6.1 Scalar Products

The scalar product of three-dimensional vectors is written as

u:v = uxvx + uyvy + uzvz;

where the dot indicates the cosine of the angle between the vectors. This

product is invariant when we rotate axes in space. When dealing with four-

vectors, we denote a scalar product by

4X�=1

A�B� = A4B4 � A1B1 � A2B2 � A3B3:

This quantity is invariant under Lorentz transformations | it is the same

in all inertial frames of reference. It is therefore often known as a (Lorentz )

scalar.

Often, theP

is omitted; using Einstein's convention of summation over

repeated indices, we can write

A�B�

which indicates that we should sum over all pairs of terms where the indices

are the same (remembering to be careful with the signs). This is just like

using u:v as shorthand for the summation over the three pairs of components

for three-vectors. Note that the scalar product of the spacetime four-vector

71

with itself gives the interval. Another convention is that four-vectors may be

written as

A = (A;A4) ;

where A represents the three spatial components. The scalar product is then

4X�=1

A�B� = A4B4: �A:B: (6.1)

In the case of the spacetime fourvector x, the scalar product x �x is of course

just the interval S2.

Note that the invariance of scalar products is extremely useful for solving

lots of relativity problems...

6.2 Energy-Momentum Four-Vector

Consider the interval

S2 = �

X�

x2

�= c

2t2 � x

2 � y2 � z

2:

In a short time �t, a particle moving with velocity u travels a distance

�x2 +�y2 +�z2 = u2�t2;

and so the element of interval between the beginning and end of that short

journey is

�S2 = c2�t2 � u

2�t2

= c2�1� �

2��t2

= c2:�t2= 2

= c2:�� 2;

where � is the proper (or Lorentz invariant) time. We saw this earlier, when

we were looking at the transformation of energy and momentum.

Now, �S is invariant, and so if we introduce a function

U� = c�x�

�S=

�x�

��;

72

it will transform in the same manner as �x�: As we make the time �t shorter

and shorter, in the usual way in calculus, we end up with

U� =dx�

d�:

The quantity U� is a four-vector, called the four-velocity because its compo-

nents are

U� = (ux; uy; uz; c) :

Notice that the scalar product of U� with itself is

U�U� = c2

�dx�

dS

�2

= �c2

(since dS2 = �P

�x2�); this is, of course, invariant under Lorentz transfor-

mations.

We can now construct the four-momentum with components

P� = mU�

P =

�p;

E

c

�;

where E is an \arbitrarily" chosen letter that represents the quantity mc2,

from the fourth component of U:

We therefore know immediately that the four-momentum will transform

in the way we have seen,

p0x

= (px � �:E=c) (6.2)

p0y

= py

p0z

= pz

E0=c = (E=c� �px) :

Furthermore, the Lorentz invariant

P� � P� =E

2

c2� p

2 = m2c2

emerges trivially. It was at this point before that we realised that it makes

sense to identify

E = mc2

73

with the total energy of the body. The four-vector notation has thus al-

lowed us to derive the Lorentz transformation for energy and momentum in

a straightforward way, without lots of tedious algebra in studying collisions.

6.3 Examples

6.3.1 Collision Threshold Energies

We earlier considered the reaction

p+ p! p+ n+ �+;

in which an incoming proton p (mass � 938 MeV/c2) of energy E0 hits a tar-

get proton at rest, to create a proton, a neutron (also of mass � 938 MeV=c2)

and a positive pion (mass 139 MeV=c2), and we derived the minimum kinetic

energy that the incident proton required for the reaction to take place. Let

us quickly do the same derivation, but using four-vectors.

The total four-momentum in the lab frame is

P =

�pp;

Ep +mpc2

c

�:

The relativistic invariant is

P � P =(Ep +mpc

2)2

c2� p

2;

which we know is the same in all frames; P � P = P0 � P 0.

Now, if we move to the centre-of-momentum frame, the four-momentum

vector (evaluated after the collision) is

P0 = (0; (mp +mn +m�)c) :

So, we have

P � P = P0 � P 0

(Ep +mpc2)

2

c2� p

2 = (mp +mn +m�)2c2:

Therefore,

(Ep +mpc2)

2

c2��E

2p�m

2pc4�

c2= (mp +mn +m�)

2c2

2mpc2Ep + 2m2

pc4 = (mp +mn +m�)

2c2

74

and, just as we found before,

Ep =(mp +mn +m�)

2

2mp

�mpc2:

This is the total energy of the incoming proton; its kinetic energy is T =

Ep � mpc2: For the masses given above, the total energy is 1226 MeV; the

kinetic energy is 288 MeV.

This looks complicated, but it isn't too bad if you work through it step

by step... it's a good example to bear in mind as you work through the

problems. Use of four-vectors doesn't make solving this particular problem

any easier, but it is useful to become familiar with the terminology so that

it can later be applied to more complex problems.

6.3.2 Matter-Antimatter annihilation

γ2

e+

γ1

e-

φ1

Figure 6.1: Annihilation of an electron and a positron to produce a pair of

gamma rays.

A positron and an electron can annihilate to produce a pair of photons.

Consider the case of a positron at rest in the laboratory and an electron

75

colliding with it, as shown in Figure 6.1. What is the energy of one of the

photons, as a function of its angle with respect to the incoming electron?

This is the kind of problem where four-vectors can really be useful; be-

cause it is no longer one-dimensional, we have several coordinates to consider

at once, and, just as with ordinary vectors in Newtonian mechanics, the use

of four-vectors here makes the algebra rather simpler than it would otherwise

be.

Conservation of the energy-momentum four-vector gives

P+ + P� = Q 1+Q 2

:

Here we have used p for the particle four-momentum, and q for the photon

four-momentum. We are not interested in photon 2, so (note this trick!) we

take its four-vector to one side, and then \square" both sides:

Q2

2= P+ � P+ + P� � P� +Q 1

�Q 1

+2P+ � P� � 2P+ �Q 1� 2P� �Q 1

:

To evaluate this, we use equation (6.1), together with the facts that

� For a photon, Q2 = E

2=c

2 � p2 = 0 (i.e., it has zero rest mass).

� The positron is at rest, so p+ = 0 and E+ = m0c2:

Therefore, we have

0 = m2

0c2 +m

2

0c2 + 0 +

2

�m0c

2

c

��E�

c

�� 2

�m0c

2

c

��E 1

c

�� 2

�E�

c

��E 1

c

�+ 2p�:q 1 :

If � is the angle between the electron and the photon,

p�:q 1 = p� � q 1 cos� = p�E 1

ccos�:

So, multiplying through by c2=2;

0 = m2

0c4 +m0c

2E� �m0c

2E 1� E� � E 1

+ p�cE 1cos �;

which may be rearranged to give

E 1

�E� +m0c

2 � p�c cos��= m0c

2�E� +m0c

2�

76

or

E 1=

m0c2 (E� +m0c

2)

(E� +m0c2)� p�c cos�

:

Thus, for a given energy E� and momentum p� of the incident electron,

the photon's energy is at a maximum if it goes forward and at a minimum

if it goes backwards. Of course, the situation is symmetric, and the same

equation will apply to either photon.

Notice the trick we used at the start of isolating the quantity that was

not of interest. If we had had both photons on the same side of the equation,

we would have had to deal with the dot product between them; as it is, the

only angle that entered was between one of the photons and the incoming

particle.

77

Chapter 7

Relativity and

Electromagnetism

7.1 Magnetic Field due to a Current

The magnetic force on a charge is

F = q (v �B) : (7.1)

But as we know, the velocity is not absolute. What happens to this force if

we move into the reference frame where the charged particle is at rest? Does

it matter which frame we are in when we measure the magnetic �eld?

Let us try applying our knowledge of relativity to the simple case of a

current-carrying wire to see what we can �nd out about the relationship

between electricity and magnetism.

We will consider a wire as a lattice of stationary positive charges, of

density �+, with negative charges of density �� moving through it at an

average velocity v� (�gure 7.1). Outside the wire, at a distance r, there is a

negative charge q�, moving with velocity v; for simplicity, we will let v = v�.

We could treat the more complicated case of an arbitrary velocity; but we

won't do so here. We shall be looking at two reference frames; in one, S, the

wire is at rest; in the second, S 0, the charge is at rest. In the S frame, there

is a magnetic force on the particle, and if it were moving freely we would see

it curve in towards the wire. But what about the S 0 frame? Clearly, since

v0 = 0, there is no magnetic force at all; does the charge, therefore, move in

towards the wire or not? And if it does, what would make it do so?

78

In the S frame, the force on the particle is given by the Lorentz equation

7.1. The magnitude of the magnetic �eld is

B =�0I

2�r:

The current I is the amount of charge passing any given point per second.

If this charge is enough to �ll a \cylinder" of the wire of length x, then

I =dq�

dt

= ��Adx

dt

= ��Av�:

So, the magnetic force on the charge has magnitude

FB = qv0�0

2�r��Av�;

and since we are considering the simple case where v = v�, we have

FB = q�0

2�r��Av

2:

Notice that, as the wire is uncharged, the positive and negative charges cancel

each other out, so we have

�� = ��+:Now, let us look at the S 0 frame. Here, the negative charges are all at rest;

but the positive charges of the lattice are moving past with speed v0+ = �v.

Firstly, we have to establish what happens to charge when we change

reference frames. The answer is: nothing at all. The total charge of a particle

remains the same, whatever its speed in our reference frame. Suppose we

take a block of metal, and heat it up; because the electrons have a di�erent

mass than the protons, their speeds will change by di�erent amounts. If the

charge on each particle depended on the speed, the charges of the electrons

and protons would no longer balance, and the block would become charged.

Likewise, the net charge of a piece of material would change in a chemical

reaction. As we have never seen any such e�ects, we have to conclude that

charge is independent of velocity.

79

But charge density is not. If we take a length L0 of wire, in which there

is a charge density �0 of stationary charges, it contains a total charge

Q0 = �0:L0:A:

If we now observe these same charges from a moving frame, they will all be

found in a piece of the material with length

L = L0

p1� v2=c2:

Therefore, since the charge Q0 and the transverse dimension A are un-

changed, we �nd

Q0 = �0:L0:A = �::AL0

p1� v2=c2;

and so

� =�0p

1� v2=c2:

Thus, the charge density of a moving distribution of charges varies in the

same way as the length or the relativistic mass of a particle.

What does all this imply for our wire? If we look at the positive charge

density in the S 0 frame, we �nd

�0+ =

�+p1� v2=c2

:

For the negative charges, though, it's the other way around. They are at rest

in S0, and moving in S; they have their \rest density" in S

0. Therefore,

�0� = ��

p1� v2=c2:

So, we �nd that the charges no longer balance | when we transfer to the

moving frame, the wire becomes electrically charged! This is why the exterior

negative charge is attracted to it. The total charge density is

�0 = �

0+ + �

0�

=�+p

1� v2=c2+ ��

p1� v2=c2:

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Since the wire is uncharged in S, we have �� = ��+, and therefore

�0 = �+

(1p

1� v2=c2� 1� v

2=c

2p1� v2=c2

)

= �+v2=c

2p1� v2=c2

:

For a charged wire, with charge density �, the electric �eld a distance r away

is

E =�:A

2�"0r:

Therefore, with our charge density �0, the force F 0 = E0q acting on the charge

in the S 0 frame is

F0 = �+

v2=c

2p1� v2=c2

A

2�"0rq:

Let us transform this force back into the S frame. We know that

F =dp

dt;

and we have seen that transverse momenta are unchanged by Lorentz trans-

formations. Therefore, we expect transverse forces to transform like the

inverse of time; in other words, the force as seen in the S frame is

F = F0p1� v2=c2

= �+v2=c

2A

2�"0rq:

Comparing this with our earlier expression for the magnetic force in S;

FB = q�0

2�r��Av

2;

we see that the magnitudes of these forces are identical provided that

c2 =

1p�0"0

;

which is Maxwell's well-known result for the speed of propagation of electro-

magnetic waves. So, the magnetic force in one reference frame is seen as a

purely electrostatic force in another.

81

It is perhaps not surprising to learn that magnetism and electricity are

not independent forces. Just as space and time are united, and mass and

energy, so electricity and magnetism are just di�erent aspects of the same

underlying electromagnetic �eld, and we have to treat both together as a

complete entity.

7.2 Lorentz Transformation of Electromagnetic

Fields

We shall not derive the Lorentz transformation of electromagnetic �elds; we

simply state the result, namely

E0k = Ek

E0? = (E+ v�B)

?

and

B0k = Bk

B0? =

�B� v� E=c2

�?;

where k and ? indicate the parallel and perpendicular components respec-

tively of the �elds. This means that a purely magnetic (or purely electric)

�eld in one frame of reference looks like a mixture of electric and magnetic

�elds in another. This should not be surprising, having seen the example of

the current-carrying wire.

7.3 Electromagnetic Waves

We have already seen how momentum and energy transform from one Lorentz

frame to another:

p0x

=

�px � �

E

c

�E0

c=

�E

c� �px

�:

For light waves, the momentum is

p = ~k (7.2)

82

where k is the wavenumber, jkj = 2�=�, and the energy carried by the wave

is

E = ~!;

where the angular frequency ! is related to the wavenumber by ! = c jkj.From this, it follows that

~k0x

= ~

�kx � �

w

c

�~!0

c= ~

�!

c� �kx

�: (7.3)

7.3.1 Phase

The phase of a (monochromatic) wave is given by

� = !t� k � r:

If we restrict ourselves to waves travelling in the x direction, we have

� = !t� kxx:

If we now use the transformations (7.3) and the familiar Lorentz transforma-

tion, we �nd that the phase �0 in the primed frame is

�0 = !

0t0 � k

0xx0

=

�!

c� �kx

� (ct� �x)�

�kx � �

w

c

� (x� �ct)

= 2�!t+ �

2kxx� kxx� �

2!t�

= 2�1� �

2�(!t� kxx)

= !t� kxx = �:

Therefore, the phase is an invariant. This should come as no surprise; the

crest of a wave, for example, is a well-de�ned object that should not depend

upon one's frame of reference. In fact, we have our argument slightly re-

versed; it is usual to begin with the invariance of the phase, and to show

that the de Broglie postulate (7.2) is consistent with relativity.

It is trivial to extend this argument to a wave travelling in three dimen-

sions.

83

7.3.2 Wave four-vector

By looking at the transformation (7.3), one can see immediately that it is

possible to construct a four-vector with components (ck;w) : The invariant is

then w2 � c

2k2 = 0 (which arises because the photon has zero rest mass).

7.3.3 Doppler Shift

The latter of equations (7.3) is of course identical with our earlier equation

for the Doppler shift:

!0 =

�w � �c � !

ccos �

�= ! (1� � cos �) :

Thus, if a source of frequency ! moves away from us at speed u = ux, we

will detect waves of frequency

!0 = ! (1� �)

= !

s1� �

1 + �:

Thus, the frequency drops, as expected for a receding source. If it moves

towards us, � ! ��, and

!0 = !

s1 + �

1� �;

again as expected.

84

r

v-=v

vq(-)

(a) Frame S; wire at rest.

I

ρ-v+=0

ρ+

r

v′- = 0

v

q(-)

(b) Frame S′; charge at rest.

I′

ρ′-

v′+ = -vρ′+

Figure 7.1: A current-carrying wire, seen in its rest frame (a) and in the rest

frame (b) of the negative charges that constitute the current.

85